Infosys HCF and LCM Questions and Answers Quiz-1

Solution: Let the three numbers be 7, 8, 9

Factors of 7 are = 1 , 7

Factors of 8 are = 1, 2, 4, 8

Factors of 9 are = 1, 3, 3, 9

We can see that the Greatest common factor is 1.

Sol: 12 = 2×2×3 = 2^2 ×3^1

15 = 3× 5

20 = 2×2×5 = 2^2×5

27 = 3×3×3 = 3^3

 

HCF = 1

LCM = 2^2 × 3^3 × 5 = 540

The product of two numbers = HCF * LCM

Let the required number is = x

So, 77 * x = 11 * 693

Hence, x=99.

Difference between numbers = 90-34 = 56, and 104- 90=14

HCF of 56 and 14 = 14

Hence, 14 is the largest number which divides the given number and leaves the same remainder in each case.

Take LCM of 9, 10, and 15. Their LCM is = 90

If the number 1936 is divided by 90, we get the remainder 46, but we need 8 as remainder.

Hence, the required number is 46 - 8 = 38.

Take LCM of 5, 6, 7, and 8. It will be = 840.

Let the number 840*x + 3 is completely divisible by 9 if the sum of digits is divisible by 9.

Let x= 2, the number will be 1683, and the sum of its digits is equals to 18 which is divisible by 9. So, the required number is 1683.

Take LCM of 3, 4, 5, 6, 7, and 8. It will be = 840.

Now, 10000/840 = 760 (remainder).

840 - 760 = 80. If we add 80 in the given number, the number is exactly divisible by 840.

So, the required number is 10080.

Note: The HCF of two prime numbers = 1

LCM/HCF =161/1 =161
161=7 * 23
The ratio of numbers: 7: 23
Hence, the expression gives15 * 7 - 7 * 23 = -56

 

Explanation:

Take LCM of 12, 15, 20, and 54. It will be = 540.

ATQ, the remainder in each case = 4.

Hence, the required number = 540+4 = 544.